Long Question (Question 10)

Question 10 (5 marks):
Solve the following simultaneous equations:
x – 3y = 1,
x² + 3xy + 9y² = 7


Solution:
$\begin{array}{l}x-3y=\mathrm{1...................}\left(1\right)\\ x^2+3xy+9y^2=\mathrm{7...................}\left(2\right)\\ \text{From }\left(1\right):x=3y+\mathrm{1...................}\left(3\right)\\ \text{Substitute }\left(3\right)\text{ into }\left(2\right):\\ {\left(3y+1\right)}^2+3\left(3y+1\right)y+9y^2=7\\ 9y^2+6y+1+9y^2+3y+9y^2-7=0\\ 27y^2+9y-6=0\\ 9y^2+3y-2=0\\ \left(3y-1\right)\left(3y+2\right)=0\\ y=\frac{1}{3}\text{ or }-\frac{2}{3}\\ \\ \text{Substitute }y\text{ into }\left(3\right):\\ \text{When }y=\frac{1}{3}\\ x=3\left(\frac{1}{3}\right)+1=2\\ \text{When }y=-\frac{2}{3}\\ x=3\left(-\frac{2}{3}\right)+1=-1\\ \\ \text{Hence, the solutions are }x=2,y=\frac{1}{3}\text{ or }x=-1,y=-\frac{2}{3}.\end{array}$